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If ABCD is a square, what are the coordinates of C?

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If ABCD is a square, what are the coordinates of C? [#permalink]

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22 Mar 2014, 15:00
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squarecoordinates_figure.PNG [ 5.13 KiB | Viewed 8245 times ]
If ABCD is a square, what are the coordinates of C?

A. $$\sqrt{3} , \sqrt{3}$$

B. $$\sqrt{3}, 1 + \sqrt{3}$$

C. $$2 \sqrt{3} , \sqrt{3}$$

D. $$1 + \sqrt{3} , \sqrt{3}$$

E. $$\sqrt{3} , 2 \sqrt{3}$$

Guys - I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 Mar 2014, 05:41, edited 1 time in total.
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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23 Mar 2014, 05:44
enigma123 wrote:

If ABCD is a square, what are the coordinates of C?

A. $$\sqrt{3} , \sqrt{3}$$

B. $$\sqrt{3}, 1 + \sqrt{3}$$

C. $$2 \sqrt{3} , \sqrt{3}$$

D. $$1 + \sqrt{3} , \sqrt{3}$$

E. $$\sqrt{3} , 2 \sqrt{3}$$

Guys - I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees?

That's because the straight line is 180 degrees (30+90+60) and not 150 degrees (30+90+30).
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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24 Mar 2014, 20:17
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In this example, all the right triangles formed would be 30 - 60 - 90 as per diagram below;
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If ABCD is a square, what are the coordinates of C? [#permalink]

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28 Jul 2014, 17:39
If Side =2 then the diagonal equals $$2\sqrt{2}$$

The diagonal equals the distance between the 2 points. So I assume we can use the distance between 2 points formula.

D=$$\sqrt{(x2-x1)^2+(y2-y1)^2}$$
$$D^2$$=$$(x2-x1)^2+(y2-y1)^2$$

$$2\sqrt{2}^2$$=$$(x2-0)^2+(y2-1)^2$$

$$2\sqrt{2}^2$$=$$(1+\sqrt{3})^2+(\sqrt{3}-1)^2$$

8=$$(1+\sqrt{3})^2+(\sqrt{3}-1)^2$$

8=4+2

???

Can anyone help and tell me where I went wrong.

I thought I could apply the distance between 2 points formula because we knew the distance, and points X1 and Y1.
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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28 Jul 2014, 20:26
If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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28 Jul 2014, 20:53
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GeorgeA023 wrote:
If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

It is a thumb rule; if a right triangle has angles 30 - 60 - 90; sides correspondingly opposite would be $$1 - \sqrt{3} - 2$$
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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28 Jul 2014, 20:55
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GeorgeA023 wrote:
If Side =2 then the diagonal equals $$2\sqrt{2}$$

The diagonal equals the distance between the 2 points. So I assume we can use the distance between 2 points formula.

D=$$\sqrt{(x2-x1)^2+(y2-y1)^2}$$
$$D^2$$=$$(x2-x1)^2+(y2-y1)^2$$

$$2\sqrt{2}^2$$=$$(x2-0)^2+(y2-1)^2$$

$$2\sqrt{2}^2$$=$$(1+\sqrt{3})^2+(\sqrt{3}-1)^2$$

8=$$(1+\sqrt{3})^2+(\sqrt{3}-1)^2$$

8=4+2

???

Can anyone help and tell me where I went wrong.

I thought I could apply the distance between 2 points formula because we knew the distance, and points X1 and Y1.

You can. Everything is correct except the last calculation.

8=$$(1+\sqrt{3})^2+(\sqrt{3}-1)^2$$

$$8= 1^2 + \sqrt{3}^2 + 2*1*\sqrt{3} + \sqrt{3}^2 + 1^2 - 2*\sqrt{3}*1$$
$$8 = 1 + 3 + 3 + 1 = 8$$
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8002 Location: Pune, India Re: If ABCD is a square, what are the coordinates of C? [#permalink] Show Tags 28 Jul 2014, 21:11 enigma123 wrote: Attachment: squarecoordinates_figure.PNG If ABCD is a square, what are the coordinates of C? A. $$\sqrt{3} , \sqrt{3}$$ B. $$\sqrt{3}, 1 + \sqrt{3}$$ C. $$2 \sqrt{3} , \sqrt{3}$$ D. $$1 + \sqrt{3} , \sqrt{3}$$ E. $$\sqrt{3} , 2 \sqrt{3}$$ There are many other ways of solving this question: One method is to note that the sides of 30-60-90 triangle are in the ratio 1:$$\sqrt{3}$$:2. Since OA has length 1, OD has length $$\sqrt{3}$$. Because the square is tilted, x coordinate of C will be more than x coordinate of D. So options (A), (B) and (E) are out. Also, the square is slightly titled so the x coordinate of C will not be twice of x coordinate of D. Hence option (C) is not possible either. Answer must be (D). Another method is graphing. It is discussed in these two posts: http://www.veritasprep.com/blog/2013/04 ... ry-part-i/ http://www.veritasprep.com/blog/2013/04 ... y-part-ii/ We see that coordinates of D are ($$\sqrt{3}$$, 0) because of the 30-60-90 triangle. Now imagine that the x axis is turned 90 degrees at D. So the y coordinate of C will be $$\sqrt{3}$$. Also since OA is of length 1, when you rotate it by 90 degrees, the x coordinate of C is 1 more than the x coordinate of C i.e. it is $$\sqrt{3} + 1$$. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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29 Jul 2014, 09:42
Karishma,

Thank you so much. This is huge help. I can not believe I missed that.
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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06 Aug 2014, 04:53
Is this a special pythagorean triangle like the ones with sides 3,4,5 and 5,12,13?
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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06 Aug 2014, 22:41
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unceldolan wrote:
Is this a special pythagorean triangle like the ones with sides 3,4,5 and 5,12,13?

Pythagorean triplets give you the ratio of sides where all three lengths of sides are integers - the most common one is 3, 4, 5.

This is the ratio of sides in case the right triangles are special triangles:

Triangle with angles 30-60-90 (occur often) - Here ratio of sides is 1: root 3: 2 (Not a pythagorean triplet since all sides are not integers)
Isosceles right triangle i.e. angles are 45-45-90 - Here ratio of sides is 1:1:root 2 (Not a pythagorean triplet since all sides are not integers)
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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18 Jul 2015, 04:18
PareshGmat wrote:
GeorgeA023 wrote:
If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

It is a thumb rule; if a right triangle has angles 30 - 60 - 90; sides correspondingly opposite would be $$1 - \sqrt{3} - 2$$

Can anyone Pls. explain , how did we get 1 for one of the sides of the blue triangle. I solved till that point and could not proceed.
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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18 Jul 2015, 04:40
robinpallickal wrote:

Can anyone Pls. explain , how did we get 1 for one of the sides of the blue triangle. I solved till that point and could not proceed.

As mentioned above, after you get the angles as 30-60-90 in the blue triangle, you need to remember that the sides for a 30-60-90 triangle are in the ratio $$1:\sqrt{3}:2$$ with '1' side opposite to 30 degree angle. Thus you get 1 for that length.

Another way is by trigonometry: in the blue triangle, you know that the hypotenuse =2

Thus in this triangle, cos (60) = base / hypotenuse ---> 1/2 = base / 2 ---> base = 1 (as cos(60) = 0.5)
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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18 Jul 2015, 05:31
enigma123 wrote:
Attachment:
squarecoordinates_figure.PNG
If ABCD is a square, what are the coordinates of C?

A. $$\sqrt{3} , \sqrt{3}$$

B. $$\sqrt{3}, 1 + \sqrt{3}$$

C. $$2 \sqrt{3} , \sqrt{3}$$

D. $$1 + \sqrt{3} , \sqrt{3}$$

E. $$\sqrt{3} , 2 \sqrt{3}$$

Guys - I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees?

An explanation using property of slopes!

Product of the slopes of Perpendicular lines = -1

$$m_1*m_2 = -1$$
Since AO = 1 therefore OD = √3 [30-60-90 Triangle ratio]
Let, Slope of Line AD, $$m_1 = -1/√3$$
i.e. Slope of Line DC, $$m_2 = √3$$
in Triangle i.e. Ratio of Y-Co-ordinate of C to Horizontal Distance of CD on X-Axis = √3/1

i.e. x Co-ordinate of C = OD+1= √3+1

i.e. Co-ordinates of C = ((√3+1),√3)

i.e. Option D
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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22 Nov 2015, 11:36
since we have 30-60-90 triangle, we can deduct that the sides of the small triangle with the base on the x axis is 1-sqrt(3)-2
which makes the length of a side of the square = 2.
since all the sides are equal, we can draw a line from point c to the x-axis to form a 30-60-90 triangle.
since we have a similar triangle with the first one, we see that the hypotenuse is the same = 2. which makes that the point on y axis is sqrt(3)

now, only points A, C, and D have y coordinate of point c sqrt(3). B and E are thus eliminated.
now, from the origin to the point D, we have a distance of sqrt(3). The distance from point D to the line perpendicular to the x-axis which form a 30-60-90 triangle is 1. Thus, the x-coordinate of the point C must be 1+sqrt(3)

Eliminate A, and C.

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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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11 Apr 2016, 21:11
PareshGmat wrote:
GeorgeA023 wrote:
If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

It is a thumb rule; if a right triangle has angles 30 - 60 - 90; sides correspondingly opposite would be $$1 - \sqrt{3} - 2$$

For Red Triangle
1/2 =1/h
h=2

cos30=\sqrt{3}/2=x/2
x axis length = \sqrt{3}

now the blue side
cos 60=1/2=x/2
x=1

so total lenght of the x axis is 1+\sqrt{3}

only D is the suitable option
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Re: If ABCD is a square, what are the coordinates of C? [#permalink]

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Re: If ABCD is a square, what are the coordinates of C?   [#permalink] 30 Nov 2017, 04:24
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